How to solve ode of this form












6












$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago


















6












$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago
















6












6








6


2



$begingroup$


$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations







share|cite|improve this question









$endgroup$




$
a_n(y')^n + a_{n-1}(y')^{n-1} + cdots + a_0 y' =0
$



I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









user47475user47475

634




634








  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago
















  • 1




    $begingroup$
    I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
    $endgroup$
    – Seth
    1 hour ago






  • 1




    $begingroup$
    Do you mean $cdots + a_1 y^prime + a_0$ at the end?
    $endgroup$
    – parsiad
    1 hour ago










  • $begingroup$
    Would make sense @parsiad
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    @NeoDarwin: encouraging OP to edit the post.
    $endgroup$
    – parsiad
    1 hour ago










1




1




$begingroup$
I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
$endgroup$
– Seth
1 hour ago




$begingroup$
I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$
$endgroup$
– Seth
1 hour ago




1




1




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
1 hour ago




$begingroup$
Do you mean $cdots + a_1 y^prime + a_0$ at the end?
$endgroup$
– parsiad
1 hour ago












$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
1 hour ago




$begingroup$
Would make sense @parsiad
$endgroup$
– Neo Darwin
1 hour ago




1




1




$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
1 hour ago






$begingroup$
@NeoDarwin: encouraging OP to edit the post.
$endgroup$
– parsiad
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Well. As pointed out in the comments by Seth.



Let us consider an example



begin{align}
(y')^2+2y'+1=0
end{align}
then it follows $y' = -1$ so $y = -t+C$.



So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago






  • 1




    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago












  • $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago



















1












$begingroup$


TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
$$
a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
$$
By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
$$
a_{n}r^{n}+cdots+a_{1}r+a_{0}
$$
has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
$$
left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
$$
By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3112254%2fhow-to-solve-ode-of-this-form%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago
















    3












    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago














    3












    3








    3





    $begingroup$

    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.






    share|cite|improve this answer









    $endgroup$



    Well. As pointed out in the comments by Seth.



    Let us consider an example



    begin{align}
    (y')^2+2y'+1=0
    end{align}
    then it follows $y' = -1$ so $y = -t+C$.



    So, in general, you will get $y'=$ const. Then $y=text{const}. t+C$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Jacky ChongJacky Chong

    18.6k21128




    18.6k21128












    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago


















    • $begingroup$
      What if it does not have real roots?
      $endgroup$
      – Neo Darwin
      1 hour ago






    • 1




      $begingroup$
      Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
      $endgroup$
      – Jacky Chong
      1 hour ago












    • $begingroup$
      Thanks. Good answer and logical.
      $endgroup$
      – Neo Darwin
      1 hour ago
















    $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago




    $begingroup$
    What if it does not have real roots?
    $endgroup$
    – Neo Darwin
    1 hour ago




    1




    1




    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago






    $begingroup$
    Then you have a complex constant. If you only want real solutions, then you can drop all the complex solutions.
    $endgroup$
    – Jacky Chong
    1 hour ago














    $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago




    $begingroup$
    Thanks. Good answer and logical.
    $endgroup$
    – Neo Darwin
    1 hour ago











    1












    $begingroup$


    TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




    Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
    $$
    a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
    $$
    By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
    $$
    a_{n}r^{n}+cdots+a_{1}r+a_{0}
    $$
    has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
    $$
    left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
    $$
    By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
    In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$


      TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




      Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
      $$
      a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
      $$
      By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
      $$
      a_{n}r^{n}+cdots+a_{1}r+a_{0}
      $$
      has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
      $$
      left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
      $$
      By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
      In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$


        TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




        Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
        $$
        By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
        $$
        a_{n}r^{n}+cdots+a_{1}r+a_{0}
        $$
        has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
        $$
        left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
        $$
        By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
        In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.






        share|cite|improve this answer











        $endgroup$




        TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.




        Let $y:[0,T)rightarrowmathbb{R}$ where $Tleqinfty$ be a continuously differentiable function satisfying the ODE
        $$
        a_{n}(y^{prime}(t))^{n}+cdots+a_{1}y^{prime}(t)+a_{0}=0.
        $$
        By the fundamental theorem of algebra, if $a_{n}neq0$, the corresponding polynomial
        $$
        a_{n}r^{n}+cdots+a_{1}r+a_{0}
        $$
        has $n$ complex roots, call them $z_{1},ldots,z_{n}$. We can re-express the ODE as
        $$
        left(y^{prime}(t)-z_{1}right)cdotsleft(y^{prime}(t)-z_{n}right)=0.
        $$
        By the above, $y^{prime}(t)=z_{k(t)}$ where $k(t)in{1,ldots,n}$ for all $t$.
        In fact, $tmapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 55 mins ago

























        answered 1 hour ago









        parsiadparsiad

        17.5k32353




        17.5k32353






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3112254%2fhow-to-solve-ode-of-this-form%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            CARDNET

            Image
            CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
            Read more

            Boot-repair Failure: Unable to locate package grub-common:i386

            Image
            0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
            Read more

            濃尾地震

            Image
            濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
            Read more